Problem: $g(x)=\dfrac{3}{x+2}$. On which intervals is the graph of $g$ concave up? Choose 1 answer: Choose 1 answer: (Choice A) A $(2,\infty)$ only (Choice B) B $(-2,\infty)$ only (Choice C) C $(-\infty,-2)$ only (Choice D) D $(0,\infty)$ only
Explanation: We can analyze the intervals where $g$ is concave up/down by looking for the intervals where its second derivative $g''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $g'$, we are analyzing $g''$. The second derivative of $g$ is $g''(x)=\dfrac{6}{(x+2)^3}$. $g''$ is never equal to $0$. $g''$ is undefined for $x=-2$. Therefore, our only point of interest is $x=-2$. Our point of interest divides the domain of $g$ (which is all numbers except for $-2$ ) into two intervals: $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $(-\infty, \llap{-}2)$ $( \llap{-}2,\infty)$ Let's evaluate $g''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g''(x)$ Verdict $(-\infty,-2)$ $x=-3$ $g''(-3)=-6<0$ $g$ is concave down $\cap$ $(-2,\infty)$ $x=-1$ $g''(-1)=6>0$ $g$ is concave up $\cup$ In conclusion, the graph of $g$ is concave up over the interval $(-2,\infty)$ only.